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ryzh [129]
3 years ago
12

Hii!! I'm struggling a little on this, could someone help me with both the questions? thanks!! (how do i give brainliest help il

l give brainliest to the best answer)

Mathematics
1 answer:
romanna [79]3 years ago
8 0

Look AT IMAGE

Step-by-step explanation:

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Help plzz I will mark brainliestttt
Radda [10]

Answer:

A

Step-by-step explanation:

3 0
3 years ago
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The magnitude and direction of two vectors are shown in the diagram. What is the magnitude of their sum? ​
VLD [36.1K]

Separate the vectors into their <em>x</em>- and <em>y</em>-components. Let <em>u</em> be the vector on the right and <em>v</em> the vector on the left, so that

<em>u</em> = 4 cos(45°) <em>x</em> + 4 sin(45°) <em>y</em>

<em>v</em> = 2 cos(135°) <em>x</em> + 2 sin(135°) <em>y</em>

where <em>x</em> and <em>y</em> denote the unit vectors in the <em>x</em> and <em>y</em> directions.

Then the sum is

<em>u</em> + <em>v</em> = (4 cos(45°) + 2 cos(135°)) <em>x</em> + (4 sin(45°) + 2 sin(135°)) <em>y</em>

and its magnitude is

||<em>u</em> + <em>v</em>|| = √((4 cos(45°) + 2 cos(135°))² + (4 sin(45°) + 2 sin(135°))²)

… = √(16 cos²(45°) + 16 cos(45°) cos(135°) + 4 cos²(135°) + 16 sin²(45°) + 16 sin(45°) sin(135°) + 4 sin²(135°))

… = √(16 (cos²(45°) + sin²(45°)) + 16 (cos(45°) cos(135°) + sin(45°) sin(135°)) + 4 (cos²(135°) + sin²(135°)))

… = √(16 + 16 cos(135° - 45°) + 4)

… = √(20 + 16 cos(90°))

… = √20 = 2√5

5 0
3 years ago
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2 years ago
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What are the original dimensions of the square piece of paper?
Inga [223]

Answer:

6root2 × 6root2

you have to use pythagoras

a²+b²=c²

c=12

12²=144

a²+b²=144

but it is a square so I'll just change a and b to x because it shows they are equal

x²+x²=144

2x²=144

x²=72

x=6root2

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2 years ago
Factor the polynomial. 6g8 – 3g4 + 9g2 urgentt
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Answer:

3g^2 (2g^6-g^2+3) the ^ symbole is means to the power

Step-by-step explanation:

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